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2x^2-20x=9900
We move all terms to the left:
2x^2-20x-(9900)=0
a = 2; b = -20; c = -9900;
Δ = b2-4ac
Δ = -202-4·2·(-9900)
Δ = 79600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{79600}=\sqrt{400*199}=\sqrt{400}*\sqrt{199}=20\sqrt{199}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{199}}{2*2}=\frac{20-20\sqrt{199}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{199}}{2*2}=\frac{20+20\sqrt{199}}{4} $
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